3
$\begingroup$

Let $R$ be a finite commutative ring with no zero-divisors. We have proven that then $R$ has a unity.

Is it then always true that $R = (y)$ when $y \neq 0$?

We have a function $ x \mapsto xy$ where $y \neq 0$ from $R \to (y)$. Then the kernel must be zero, thus making it injective, and a bijection since both are finite. Does this always hold?

$\endgroup$
  • 1
    $\begingroup$ Yes. In other words, $R$ is a field, because in a commutative ring with unity, $y$ is invertible iff $(y) = R$. $\endgroup$ – Andreas Caranti Jan 3 '17 at 20:31
3
$\begingroup$

In the finite case you have that a commutative ring $R$ (not the zero-ring) is a field $\iff$ it has no zero-divisors (except 0 itself, depending on the Def. of zero-divisor).

Therefore the units of R are $R^{*} = R\backslash \{0\}$, hence $R=\langle y\rangle$ $ \forall y \ne 0$.

Most important part here is that it is finite. If it's not, consider $R[X]$ as a counterexample ($y = X$, $R[X]\ne \langle X\rangle $).

$\endgroup$
  • 1
    $\begingroup$ Thank you for the counter example $\endgroup$ – ZirconCode Jan 3 '17 at 21:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.